Tìm GTNN của các đa thức
a) A = x2 - 10x + 25
b) B = x2 + y2 - x + 6y + 10
c) C = 2x2 - 6x + 10
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TC
2
9 tháng 8 2016
a) \(P=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\)
\(MinP=4\Leftrightarrow x-1=0\Rightarrow x=1\)
b) \(Q=2x^2-6x\)
\(=2\left(x^2-3x\right)\)
\(=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right)\)
\(=2\left(\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right)\)
\(=-\frac{9}{2}-2\left(x-\frac{3}{2}\right)^2\le\frac{-9}{2}\)
\(MinQ=\frac{-9}{2}\Leftrightarrow x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
9 tháng 8 2016
M=x^2+y^2-x+6y+10
M=(x^2-x+1/4)+(y^2+6y+9)+3/4
M=(x-1/2)^2+(y+3)^2+3/4
\(minM=\frac{3}{4}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)
23 tháng 10 2021
a: ta có: \(P=x^2+10x+27\)
\(=x^2+10x+25+2\)
\(=\left(x+5\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi x=-5
19 tháng 10 2017
a)\(A=\left(x-5\right)^2\ge0\)
\(\Rightarrow Min=0\)dấu \(=\)xảy ra khi \(x=5\)
23 tháng 7 2018
a) \(A=x^2-10x+25\)
\(A=\left(x^2-10x+25\right)+0\)
\(A=\left(x-5\right)^2+0\)
Mà \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow A\ge0\)
Dấu "=" xảy ra khi : \(x-5=0\Leftrightarrow x=5\)
Vậy ...
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 12 2020
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
19 tháng 12 2020
A= -x2+2x+3
=>A= -(x2-2x+3)
=>A= -(x2-2.x.1+1+3-1)
=>A=-[(x-1)2+2]
=>A= -(x+1)2-2
Vì -(x+1)2 ≤0=> A≤-2
Dấu "=" xảy ra khi
-(x+1)2=0 => x=-1
Vây A lớn nhất= -2 khi x= -1
19 tháng 12 2020
B=x2-2x+4y2-4y+8
=> B= (x2-2x+1)+(4y2-4y+1)+6
=> B=(x-1)2+(2y+1)2+6
=> B lớn nhất=6 khi x=1 và y=-1/2
26 tháng 12 2021
d: \(=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
\(\text{a) }A=x^2-10x+25\\ A=x^2-2\cdot x\cdot5+5^2\\ A=\left(x-5\right)^2\\ Do\text{ }\left(x-5\right)^2\ge0\forall x\\ \Leftrightarrow A\ge0\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-5\right)^2=0\\ \Leftrightarrow x-5=0\\ \Leftrightarrow x=5\\ \text{Vậy }A_{\left(Min\right)}=0\text{ }khi\text{ }x=5\)
\(\text{b) }B=x^2+y^2-x+6y+10\\ B=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\\ B=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\\ Do\text{ }\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\ \left(y+3\right)^2\ge0\forall y\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x;y\\ \text{Dấu "=" xảy ra khi: }\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\\\left(y+3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\\ \text{ Vậy }B_{\left(Min\right)}=\dfrac{3}{4}\text{ }khi\text{ }x=\dfrac{1}{2};y=-3\)
\(\text{c) }C=2x^2-6x+10\\ C=\left(2x^2-6x+\dfrac{9}{2}\right)+\dfrac{11}{2}\\ C=2\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{11}{2}\\ C=2\left[x^2-2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{11}{2}\\ C=2\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{2}\\ Do\text{ }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{2}\ge\dfrac{11}{2}\\ \text{Dấu "=" xảy ra khi: }\\ \left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\\ \text{Vậy }C_{\left(Min\right)}=\dfrac{11}{2}khi\text{ }x=\dfrac{3}{2}\)
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b)
\(B=x^2+y^2-x+6y+10=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\left(10-9-\dfrac{1}{4}\right)\)\(B=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)